"""
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
"""


# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        self.res = False

        def bfs(root, target):
            if self.res is True:
                return
            if root is None:
                return
            if root.left is None and root.right is None and target == root.val:
                self.res = True
                return
            bfs(root.left, target - root.val)
            bfs(root.right, target - root.val)
            return

        if root is None:
            return False
        bfs(root, sum)
        return self.res
